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Entropy Change in Heat Exchange

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An ideal gas C_{P} = 7R/2 is heated in a steady flow heat exchanger from 70^{o}C to 190^{o}C, by another stream of the same ideal gas entering at 320^{o}C. The flow rates of the two streams are the same.

(i) Calculate DS of the two gas streams for countercurrent flow

(ii) What is DS_{total}?

(iii) If heating stream enters at 200^{o}C, what is DS_{total}?

Calculations:

Basis: 1 mole of gas

Entropy change for the constant pressure heating/cooling process:

DS = nC_{P} ln (T_{2}/T_{1})

Where T_{1} and T_{2} are respectively initial and final temperatures.

(i)* Entropy change of the cold gas stream:*

*
*T_{1} = 273 + 70 = 343 K

T_{2} = 273 + 190 = 463 K

n = 1

Therefore, DS = (7R/2) x ln (463/343) = 1.05 R

Entropy change of the Hot gas stream:

T_{1} = 273 + 320 = 593 K

Since the flow rates of the two streams are same,

DT_{hotstream} = DT_{coldstream} = 190 - 70 = 120 K

T_{2} = T_{1} - 120 = 593 - 120 = 473 K

n = 1

Therefore, DS = (7R/2) x ln (473/593) = -0.79 R

(ii) DS_{total} = 1.05 R + (-0.79R) = 0.26 R

(iii) *Heating stream inlet temperature = 200*^{o}C

*
*Therefore, outlet temperature of heating stream = 200 - 120 = 80^{o}C

DS for this stream = (7R/2) x ln (353/473) = -1.024 R

DS_{total} = 1.05 R + (-1.024 R) = 0.026 R

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Last Modified on: 11-Sep-2014

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